Atomic Structure Theory - Definition of Atomic Structure

Atomic Structure Theory - Definition of Atomic Structure
Atomic Structure Theory is a basic unit of matter consisting of a nucleus and its negatively charged electron cloud surrounding it. The nucleus contains a mix of positively charged protons and electrically neutral neutrons (except in Hydrogen-1 which has no neutrons). The electrons in an atom bound to the nucleus by the electromagnetic force. Similarly, a collection of atoms can bind to each other to form a molecule. Atoms containing the number of protons and electrons of the same neutral, while containing the number of protons and electrons of different positive or negative and is ion. Atoms are grouped based on the number of protons and neutrons in the atomic nucleus.
The number of protons in an atom determines the chemical element the atom, and the number of neutrons determine the isotope of the element. The term atom comes from the Greek, which means it can not be cut or something that can not be divided again. The concept of the atom as a component that can not be divided again was first proposed by the philosophers of India and Greece. In the 17th century and into the 18th, the chemists laid the foundations of this idea by showing that certain substances can not be broken down further using chemical methods. During the late nineteenth century and early twentieth century, physicists have managed to find the structure and subatomic components inside the atom, to prove that the 'atom' is not never be divided again. The principles of quantum mechanics used by physicists then successfully model the atom.
Relative to daily observations, the atoms are very small objects with masses as minor anyway. Atoms can only be monitored using special equipment such as tunneling microscope Microscopy. More than 99.9% of the mass of the atom is concentrated in the nucleus, the proton and neutron are almost the same mass. Each element has at least one isotope with unstable nuclei that can undergo radioactive decay. 

HISTORICAL OUTLINE
of the Atomic Theory and the Structure of the Atom

Development of the Atomic Theory

• Democritus (460-370 BC) First proposed the existence of an ultimate particle.  Used the word "atomos" to describe this particle. 
  
  Democritus

• Aristotle (384-322 BC) was a proponent of the continuum.  He believed in the four elements of air, earth, water and fire.  Aristotle felt that regardless of the number of times you cut a form of matter in half, you would always have a smaller piece of that matter.  This view held sway for 2000 years primarily because Aristotle was the tutor of Alexander the Great.
  
  Aristotle

• Johann Becher (1635-1682) and Georg Stahl (1660-1734) developed the Phlogiston theory which dominated chemistry between 1670 and 1790.  Basically, when something burned, it lost phlogiston to the air (after all, you could see the phlogiston leaving)  A problem with the theory was that burning of metals resulted in an increase in the mass.  This problem was solved by assigning negative mass to phlogiston.
   

• Joseph Priestly (1733-1804) discovered oxygen (which he called "dephlogisticated air") in 1774.  Priestly was an ardent phlogistonist until his dying day.  Priestly was also an early anti-war activist who favored both the American and French Revolutions.  He was shipped to the U.S. in 1791 where he lived a quiet life in Pennsylvania.  His house was used as a starting point for the American Chemical Society in 1876.  The Priestly Medal is the highest award given by to an American chemist by the Society.
  
  Joseph Priestly

• Antoine Lavoisier (1743-1794) was the first person to make good use of the balance.  He was an excellent experimenter.  After a visit with Priestly in 1774, he began careful study of the burning process.  He proposed the Combustion Theory which was based on sound mass measurements.  He named oxygen.  He also proposed the Law of Conversation of Mass which represents the beginning of modern chemistry.  To support his work, Lavoisier was associated with a tax-collecting firm and was married to the daughter of the one of the firm's executives.  Some people believe that Madame Lavoisier was every bit as good a scientist as her husband.   Unfortunately, this relationship with the tax firm led to Lavoisier's beheading at the guillotine in 1794.
  
  Marie and Antoine Lavoisier
  
  

• Joseph Proust (1754-1826) proposed the the Law of Constant Composition in 1799.  This law was very radical at the time and was hotly contested by Claude Berthollet (1748-1822).
  
  Joseph Proust

• John Dalton (1776-1844) proposed the Law of Multiple Proportions.  This law led directly to the proposal of the Atomic Theory in 1803.  He also developed the concept of the mole and proposed a system of symbols to represent atoms of different elements.  (The symbols currently used were developed by J.J. Berzelius(1779-1848)).  Dalton recognized the existence of atoms of elements and that compounds formed from the union of these atoms.  He therefore assumed that simplest ratios would be used in nature and came up with a formula for water of HO.  He then assigned a relative atomic weight of one to hydrogen and developed a relative atomic weight scale from percent composition data and assumed atomic ratios.  Today we would refer to these as equivalent masses.  John Dalton also discovered color blindness, an affliction from which he suffered.  He determined that five percent of the male population and less than one-tenth percent of the female population was color blind.
  
  John Dalton

• Joseph Gay-Lussac ( 1778-1850) announced the Law of Combining Volumes in 1808.  He showed that at the same temperature and pressure, two volumes of hydrogen gas reacted with one volume of oxygen gas to produce two volumes of water (as a gas).
   
  Joseph Gay-Lussac

• Amadeo Avogadro (1776-1856) proposed what is now known as Avogadro's Hypothesis in 1811.  The hypothesis states that at the same temperature and pressure, equal volumes of gases contain the same number of molecules or atoms.  When this is combined with Gay-Lussac's Law of Combining Volumes, the only possible formulas for hydrogen, oxygen and water are H₂, O₂ and H₂O, respectively.  The solution to the atomic weight problem was at hand in 1811.  However, Avogadro's Hypothesis was a radical statement at the time and was not widely accepted until fifty years later.
  
  Amedeo Avogadro

• Stanislao Cannizzaro (1826-1910), in 1860 at the Karlsruhe Conference, proposed that Avogadro's Hypothesis be accepted and the implications used for a period of five years.  At the end of this five year period, a new conference would be called to discuss any problems that might develop; this second conference was never called.
  
  
  
  
   
  
  
  
  
  
  
  
   
  Stanislao Cannizzaro

• Dimitri Mendeleev (1834-1907) proposed the periodic law and developed the first periodic table in 1869.  Medeleev's table was arranged according to increasing atomic weight and left holes for elements that were yet to be discovered.
  
  
  
  
  
  
  
  
  
  
  
  Dimitri Mendeleev

 

Development of Atomic Structure

 

• J. J. Thomson (1856-1940) identified the negatively charged electron in the cathode ray tube in 1897.  He deduced that the electron was a component of all matter and calculated the charge to mass ratio for the electron.
                 e/m = -1.76 x 10⁸ coulombs/g
  Thomson and others also studied the positive rays in the cathode ray tube and discovered that the charge to mass ratio depended on filling gas in the tube.  The largest charge to mass ratio (smallest mass) occurred when hydrogen was the filling gas.  This particle was later identified as the proton.
                 e/m =  +9.58 x 10⁴ coulombs/g
  Thomson proposed the "plum pudding" model of the atom.  In this model, the volume of the atom is composed primarily of the more massive (thus larger) positive portion (the plum pudding).  The smaller electrons (actually, raisins in the plum pudding ) are dispersed throughout the positive mass to maintain charge neutrality.
  
  
  
  
  
  
  
  
   
  
  
  Joseph John Thomson

 

• Robert Millikan (1868-1953) determined the unit charge of the electron in 1909 with his oil drop experiment at the University of Chicago.  Thus allowing for the calculation of the mass of the electron and the positively charged atoms.
                 e = 1.60 x 10^-19 coulombs
  
  
  
  
  
  
  
  
  
   
  
  Robert Millikan

 

• Ernst Rutherford (1871-1937) proposed the nuclear atom as the result of the gold-foil experiment in 1911.  Rutherford proposed that all of the positive charge and all of the mass of the atom occupied a small volume at the center of the atom and that most of the volume of the atom was empty space occupied by the electrons.  This was a very radical proposal that flew in the face of Newtonian Physics.  Although positive particles had been discussed for some time, it was Rutherford in 1920 that first referred to the hydrogen nucleus as a proton.  Also in 1920, Rutherford proposed the existence of the third atomic particle, the neutron.
   
  Ernst Rutherford

• Henry Moseley (1887-1915) discovered that the energy of x-rays emitted by the elements increased in a linear fashion with each successive element in the periodic table.  In 1913, he proposed that the relationship was a function of the positive charge on the nucleus.  This rearranged the periodic table by using the atomic number instead of atomic mass to represent the progression of the elements.  This new table left additional holes for elements that would soon be discovered.  Unfortunately, Moseley was killed at Gallipoli during WWI.
  
  Henry Moseley

• Francis Aston (1877-1945) invented the mass spectrograph in 1920.  He was the first person to observe isotopes.  For example he observed that there were three different kinds of hydrogen atoms.  While most of the atoms had a mass number of 1 he also observed hydrogen atoms with mass numbers of 2 and 3.  Modern atomic masses are based on mass spectral analysis.  His work led Rutherford to predict the existence of the neutron.
  
  Francis Aston

 

James Chadwick (1891-1974) discovered the neutron in 1932.  Chadwick was a collaborator of Rutherford's.  Interestingly, the discovery of the neutron led directly to the discovery of fission and ultimately to the atomic bomb. 

What is Atomic Number and Atomic Weight?

• Atomic number of an element is the number of protons in the nucleus of an atom. Since atoms are electrically neutral, the number of protons equal the number of electrons in an atom.
• Atomic weight (or relative atomic mass) of an element is the number of times an atom of that element is heavier than an atom of hydrogen. The atomic weight of hydrogen is taken to be unity [1].
• Mass number of an element is the sum of the number of protons and neutrons in the nucleus of an atom.

Enthalpy and enthalpy changes Enthalpy and enthalpy change (ΔH)

Enthalpy and enthalpy changesEnthalpy and enthalpy change (ΔH)
Enthalpy (H) is the amount of energy that the system at a constant pressure. Enthalpy (H) is defined as the amount of energy contained in the system (E) and work (W).H = E + Wwith:W = P × VE = energy (joules)W = work system (joules)V = volume (liters)P = pressure (atm)The law of conservation of energy explains that energy can not be created and can not be destroyed, but can only be converted from one form of energy into another form of energy. Energy value of the material can not be measured, which can be measured is the change in energy (ΔE). Similarly, enthalpy, enthalpy can not be measured, we can only measure the change in enthalpy (ΔH).ΔH = Hp - Hrwith:ΔH = change in enthalpyHp = enthalpy of productsHr = enthalpy of the reactants or reagentsa. If the product H> H reactants, then ΔH is positive, it means that the absorption of heat from the environment to the system.b. When the reactant H> H products, then ΔH is negative, which means that the release of heat from the system to the environment.
Mathematically, enthalpy change (ΔH) can be derived as follows.H = E + W (1)At constant pressure:ΔH = ΔE + PΔV (2)ΔE = q + W (3)Wsistem =-PV (4)Substitution of equation (3) and (4) in equation (2):H = (q + W) + PΔVH = (q - PΔV) + PΔVH = qThus, at constant pressure, the change in enthalpy (ΔH) is equal to the heat (q) absorbed or released (James E. Brady, 1990).A wide range of chemical reactions based on the heat released / absorbed heat (Martin S. Silberberg, 2000):a. Chemical reactions that require or absorb heat are called endothermic reactions.Example:Termination of the reaction bonding in the molecule H2 elements are:H2 → 2 H + a = ΔH kJEndothermic with ΔH is positive (+).b. The chemical reaction that releases heat is called exothermic reactions.Example:Bond formation reactions in the H2 molecule elements are:2H → H2 kJ ΔH =-aExothermic reaction with ΔH marked (-).Diagram enthalpy (energy level diagram)

 Enthalpy change for the combustion of 1 mole of a substance is measured at 298 K, 1 atm is called the standard enthalpy of combustion (standard enthalpy of combustion), expressed by ΔHc0. Enthalpy of combustion is also expressed in kJ mol -1.
Price enthalpy of combustion of various types of materials at 298 K, 1 atm are given in Table 3 below.
Table 3. Enthalpy of combustion of various types of materials at 298 K, 1 atmgb18
Burning gasoline is exothermic process. If the gas thought to be composed of isooctane, C8H18 (gasoline components) to determine the amount of heat released in the combustion of 1 liter of gasoline. Given the enthalpy of combustion of isooctane = -5460 kJ mol-1 and density isooktan = 0.7 kg L -1 (H = 1 and C = 12).
Answer:Enthalpy of combustion of isooctane is - 5460 kJ mol-1. The mass of 1 liter of gasoline = 1 liter x 0.7 kg L-1 = 0.7 kg = 700 grams. Isooctane = 700 g mol gram/114 mol-1 = 6.14 mol. So the heat released in the combustion of 1 liter of gasoline is: 6:14 x 5460 kJ mol = 33524.4 kJ mol -1.Enthalpy of DecompositionDecomposition reaction is the reverse of the formation reaction. Therefore, in accordance with the principle of conservation of energy, equal to the value of the enthalpy of formation enthalpy of decomposition, but opposite in sign.



 Determination of Enthalpy Changes and Hess's Law
To determine the enthalpy changes in chemical reactions commonly used tools such as the calorimeter, thermometer and so on that may be more sensitive.
Calculation: DH reaction = S DHfo products - S reactants DHfo

Hess Law
"The amount of heat required or released in a chemical reaction does not depend on the course of the reaction but is determined by the initial state and the end."
Example: C (s) + O2 (g) ® CO2 (g), DH = x kJ ® 1 stageC (s) + 1/2 02 (g) ® CO (g), DH = y kJ ® 2 stagesCO (g) + 1/2 O2 (g) ® CO2 (g) DH = z kJ-------------------------------------------------- ---------- +C (s) + O2 (g) ® CO2 (g) DH = y + z kJ
According to Hess's Law: x = y + z



Reaction enthalpy change in release or absorbed only depends on the initial state and the final state. The higher the reaction temperature the faster the rate of reaction.
           Changes in heat on a substance or perubhan system is determined by the temperature, time and substance specific heat, specific heat is the heat required to raise the temperature of 1 gram of a substance as high as 1 k.
           Calculating the amount of heat released or absorbed berdsarkan temperature in solution time and turns the heat capacity of materials specified calories.
Enthalpy is a term in thermodynamics that states the amount of internal energy of a thermodynamic system plus the energy used to do work. Enthalpy can not be measured, which can be calculated is the value changes. Mathematically, the enthalpy change can be formulated as follows:
ΔH = ΔU + PΔV
where:
• H = enthalpy of the system (joules)
• U = internal energy (joules)
• P = pressure of the system (Pa)
• V = volume of the system ()

CHEMICAL REACTION IN ELECTROLYTE SOLUTIONS SOLUTIONS STIKIOMETRI

CHEMICAL REACTION IN ELECTROLYTE SOLUTIONS SOLUTIONS STIKIOMETRI 

Acid-base reaction (neutralization reaction).Acid-base reaction or neutralization reaction is the reaction between the acid (H +) and a base (OH-) produced a neutral H2O.Adapaun example neutralization reaction is as follows:
1. Reactions: Acid + base -> salt + waterHNO3 (aq) + KOH (aq) -> KNO3 (aq) + H2O (l)H2SO4 (aq) + Ca (OH) 2 (aq) -> CaSO4 (aq) + H2O (l)
 2. Reaction: Oxide + Acid Tongue -> Salt + Water2HCl (aq) + CaO (s) -> CaCl2 (aq) + H2O (l)2HNO3 (aq) + Na2O (s) -> Na (NO3) 2 (aq) + H2O (l)
 3. Reactions: Ammonia + Acid -> SaltHCl (aq) + NH3 (g) -> NH4Cl (aq)H2SO4 (aq) +2 NH3 (g) -> (NH4) 2SO4 (aq)Ammonia (NH3), including bases that form molecular compounds that are distinguished from the other two types of bases, ie, ionic compounds that can release OH-ions and okisda bases. There are other basic molecular compounds such as methylamine (CH3NH2) but the reaction is not common as ammonia.4. Reaction: Oxides of acid + base -> salt + waterSO3 (g) + 2NaOH (aq) -> Na2SO4 (aq) + H2O (l)CO2 (g) + Mg (OH) 2 (aq) -> MgCO3 (aq) + H2O (l)Reaction crowding-MetalCrowding-metal reaction is a reaction in which the metal pressing of other metal cations or hydrogen in a compound. This reaction can take place if the metals are on the left of the metal / H which urged the Volta series. In this reaction, the reaction product in the form of metal deposition, gas, and water.Volta Series is a sequence of elements whose reduction potential is based on the data. The following are some elements that can be memorized by a potential reduction sequence:Li - K - Ba - Ca - Na - Mg - Al - Mn - Zn - Fe - Ni - Sn - Pb - (H) - Cu - Hg - Ag - Pt - AuAs an example of crowding-metal reactions are as follows:1. Reactions: 1 + Metal Salt 1 -> 2 + Metal Salt 2Zn (s) + CuSO4 (aq) -> ZnSO4 (aq) + Cu (s)2AL (s) + 3FeSO4 (aq) -> Al2 (SO4) 3 (aq) + 3Fe (s)Cu (s) + Na2SO4 (aq) -> no reaction because Cu is on the right row voltaic2. Reaction: Metal + Acid -> Salt + Hydrogen GasMg (s) + HCl (aq) -> MgCl2 (aq) + H2 (g)Zn (s) + H2SO4 (aq) -> ZnSO4 (aq) + H2 (g)Ag (s) + HCl (aq) -> Ag does not react due to the right of the voltaic series3. Reaction: Metal + Acid -> Salt + Water + Gas2Fe (s) + 6 H2SO4 (aq) -> Fe2 (SO4) 3 (aq) + 6 H2O (l) + 3SO2 (g)Cu (s) + 4HNO3 (aq) -> Cu (NO3) 2 (aq) + 2H2O (l) + 2NO2 (g)Metathesis reactions (Exchange Pair)Metathesis reaction is the reaction of the ion pair exchange two electrolytes.AB + CD -> AC + BDIn this reaction is at least one reaction product will form a precipitate, a gas, or a weak electrolyte. Gas can be derived from the decomposition of hypothetical substances (acids and bases hypothetical break down into gas and water) that are not as stable as the following:H2CO3 -> CO2 (g) + H2O (l)H2SO3 -> SO2 (g) + H2O (l)NH4OH -> NH3 (g) + H2O (l)
 As an example of metathesis reactions (exchange partner) are as follows:1. Reaction: Acid salt 1 + 1 -> 2 + Acid Salt 2AgNO3 (aq) + HBr (aq) -> AgBr (aq) + HNO3 (aq)ZnS (s) + 2HCl (aq) -> ZnCl2 (aq) + H2S (aq)
 2. Reaction: Salt 1 + Bases 1 -> Salt 2 + 2 BasesCuSO4 (aq) + 2NaOH (aq) -> Na2SO4 (aq) + Cu (OH) 2 (aq)NH4Cl (aq) + KOH (aq) -> KCl (aq) + NH4OH (aq)
 3. Reaction: Salt Salt 1 + 2 -> 3 + Garam Garam 4AgNO3 (aq) + NaCl (aq) -> AgCl (aq) + NaNO3 (aq)2KNO3 (aq) + MgCl2 (aq) -> 2KCl (aq) + Mg (NO3) 2 (aq)

Stoichiometric

Stoichiometric is the study of quantitative relationships of substances in a reaction stoichiometric kimia.Dari literature means measuring elements. This term is generally used more widely, which includes an assortment pengukurab broader and includes calculations of chemical substances and mixtures.Directors stoichiometric reaction is perfect, can be calculated and measured, and does not leave the rest of the reaction (reaction low), when reacted not produce residual mole. Examples of stoichiometric reaction mixture systems NaOH - HCl is 4 ml NaOH and 4 ml HCl, 5 ml and 2.5 ml H2SO4 NaOH.Non-stoichiometric reactions are corrections that leave the rest of the overall reaction or no reaction react exhausted (no limiting reagent).The maximum point is a stoichiometric point shows the highest number of a graph between the measured properties of the quantity pereaksinya.Contoh of the maximum point on the stoichiometric system experiment was conducted 3 times with different volumes are 2ml H2SO4, 4 ml, 6ml and HCl as 6ml, 4ml, 2ml from the mixing of the two compounds produced the same temperature of the mixture that is 32 so it can not be determined maximum temperature and minimum temperature.Point or minimum point is the lowest temperature in a reaction and a point that shows the number pereaksinya quantity.Limiting reagent is the reaction of (a substance) are depleted react after experiencing a reaction process because it limits the possibility of continued reaction. Limiting reagent is determined by comparing the number of moles of each reagent according to the coefficient of the reaction and then select the value of the reagent for the number of moles of reactants and reagents terkecil.Contoh coefficient of limiting reactant in the stoichiometric system NaOH - HCl is 2 ml NaOH + 6 ml of HCl that a reagent limiting NaOH, 6 ml 2 ml HCl + NaOH the limiting reagent HCl, 2 ml NaOH + 6 ml H2SO4 is the limiting reagent 2NaOH, 6 ml NaOH + 2 ml H2SO4 is the limiting reagent H2SO4.At the stoichiometric NaOH and HCl systems, some use stoichiometric reactions and non-stoichiometric reactions. The reaction is stoichiometric reaction experiments showed 4 ml 1 M NaOH and 4 mL of 1 M HCl for pereaksinya no trace. While the reaction that shows the reaction is non-stoichiometric experiments 2 ml 1 M NaOH and 6 ml of HCl 1 M, 6 mL of NaOH and 2 ml of HCl, because pereaksinya still leave the rest.Stoichiometric chemical calculations it!
Is essentially the concept of the mole and its relationship with:
mass = moles x atomic mass or relative molecular
volume (STP) = moles x 22.4 liters / mol
pressure = (moles x 0082 L.atm / mol.K x temperature) / volume
number of particles = mol x 6:02 x 10 ^ 23
concentration = moles / volume of solution
etc..

 Stochiometric is fundamental in the study of chemistry, particularly concerning the quantity of the substance involved both before and after reaksi.Stokiometri is a comparison of the relative reaction to the formation of compounds or substances and elements. The fundamental concept is the concept of stoichiometric mole. From the concept of the mole, Stokiomeri can be used for molecular weight measurement, weight and weight of equivalent elements. Of moles can be known how many atoms or molecules are present in a solution or some volume of a gas at a certain temperature. From the concept of the mole can also be written chemical equation that occurs between two objects element or compound.A chemical equation is clearly summarize the information about the substances involved in the corrections. This equation is not simply a qualitative statement describing how much of the reactants and reaction products involved.There is one very easy way to learn some stoichiometric reactions, namely the continuous method. By varying the molar quantities, pereaksinya be changed while kuantias molar fixed, then observed his physical one. One of the particular physical properties dipikih to be examined, such as Mass, time, volume, temperature, or absorption. Therefore kuntitas pereaksinya different, then the price perubaha physical properties of this system can be used to menghitun stoichiometric system.The experiment itself is done in order to understand the concept of stoichiometric practitioner in that practitioner langsung.Selain can also determine the reaction is stoichiometric or non stokioketri and can also determine the maximum and minimum points on a stoichiometric system.
 Stoichiometric or count the ways chemistry is calculation oriented fundamental laws of chemistry. Be aware with the size of the material being studied in chemistry is so very small, so there is a special unit for the amount or concentration of a substance. Besides, an element has a period of relative atomic (Ar) nor typical relative molecular mass (Mr) of a compound and all of this related to the concept of the mole.

	Examples of questions:1. What percentage of the levels of calcium (Ca) in calcium carbonate? (Ar: C = 12; O = 16; Ca = 40)Answer:1 mol CaCO, containing 1 mol Ca + 1 + 3 mol C mol OMr CaCO3 = 40 + 12 + 48 = 100So the level of calcium in CaCO3 = 40/100 x 100% = 40% 2. A total of 5.4 grams of aluminum metal (Ar = 27) was treated with excess dilute hydrochloric acid according to the reaction:2 Al (s) + 6 HCl (aq)    2 AlCl3 (aq) + 3 H2 (g)How many grams of aluminum chloride and how many liters of hydrogen gas produced at standard conditions? Answer:Of the equation can be expressed2 mol Al x 2 mol AlCl3   3 mol H25.4 grams of Al = 05/04/27 = 0.2 molSo:AlCl3 formed = 0.2 x Mr AlCl3 = 0.2 x 133.5 = 26.7 gramsThe volume of H2 gas produced (0o C, 1 atm) = 3/2 x 0.2 x 4.22 = 6.72 liters 3. An iron ore contains 80% Fe2O3 (Ar: Fe = 56, O = 16). Oxide is reduced with CO gas to produce iron.How many tons of iron ore required to make 224 tons of iron?Answer:1 mol Fe2O3 contains 2 mol Fethen: mass Fe2O3 = (Mr Fe2O3 / 2 Ar Fe) Fe = mass x (160/112) x 224 = 320 tonsSo the iron ore required = (100/80) x 320 tons = 400 tons 4. To determine the crystal water 24.95 grams of copper sulfate salt crystal is heated until all the water evaporates. After heating the salt mass to be 15.95 grams. How much water is contained in the crystal salt?Answer:suppose formula is CuSO4 salt. xH2OCuSO4.  xH2O CuSO4 + xH2O24.95 grams of CuSO4. xH2O + 18x = 159.5 moles15.95 grams CuSO4 = 159.5 mol = 0.1 molaccording to the above equation can be stated that:number of moles of CuS04. CuSO4 xH2O = mol; thus equation24.95 / (159.5 + 18x) = 0.1  x = 5So the formula is CuS04 salt. 5H2O Empirical Formula and Molecular Formula The empirical formula is the simplest formula of a compound.This formula simply states ratio of the number of atoms contained in the molecule.The empirical formula of a compound can be determined if known one:- Ar mass and each element-% Ar mass and each element- Ar mass ratio and each element Molecular formula: if the empirical formula is known and it is also known to Mr formulas can be determined.Example: A compound C den H contains 6 grams of C and 1 gram H.Determine the empirical formula and the molecular formula of the compound where it is known Mr = 28!Answer: moles C: moles H = 6/12: 1/1 = 1/2: 1 = 1: 2So the empirical formula: (CH2) nWhen Mr compound = 28 then: 12n + 2n = 28   14N = 28 n = 2So the molecular formula: (CH 2) 2 = C2H4Example: For a 20 ml oxidize hydrocarbons (CxHy) required oxygen in a gaseous state as much as 100 ml and 60 ml of CO2 produced. Determine the molecular formula of the hydrocarbon?Answer: hydrocarbon combustion equation in generalCxHy (g) + (x + 1/4 y) O2 (g)    x CO2 (g) + 1/2 y H2O (l)The coefficient indicates the reaction mole ratio of the substances involved in the reaction.According to Gay Lussac gases on p, t the same time, the number of moles is proportional to the volumeThen:CxHy mol: mol O2: mol CO2 = 1: (x + 1/4y): x20: 100: 60 = 1: (x + 1/4y): x1: 5: 3 = 1: (x + 1/4y): xor:1: 3 = 1: x    x = 31: 5 = 1: (x + 1/4y)    y = 8So hydrocarbon formula is: C3H8

CHEMICAL REACTION

Chemical Reactions

        Now that we know the how and why of chemical bonding, we can look at some chemical reactions.  Chemical reactions happen all around us: when we light a match, start a car, eat dinner or walk the dog.  A chemical reaction is the pathway by which two substances bond together.  In fact we have already discussed several chemical reactions.  One we have mentioned is the reaction of hydrogen with oxygen to form water.  To write the chemical reaction you would place the reactants (the substances reacting) on the left with an arrow pointing to the the products (the substances being formed).  Given this information, one might guess that the reaction to form water is written: H + O  H₂O However there are 2 problems with this chemical reaction.  First, because atoms like to have full valence shells, single H or O atoms are rare (and unhappy) creatures.  As we saw in the previous lesson, both hydrogen and oxygen react with themselves to form the molecules H₂ and O₂, respectively.  These hydrogen and oxygen molecules are much more common.  Given this correction, one might guess that the reaction looks like this:
H₂ + O₂ H₂O But we still have one problem.  As written, this equation tells us that 1 hydrogen molecule (with 2 H atoms) reacts with 1 oxygen molecule (with 2 O atoms) to form 1 water molecule (with 2 H atoms and 1 O atom).  In other words, we seem to have lost 1 O atom along the way!  To write a chemical reaction correctly, the number of atoms on the left side of a chemical equation has to be precisely balanced with the atoms on the right side of the equation.  How does this happen in our example?  In actuality, the O atom that we 'lost' reacts with a 2nd molecule of hydrogen to form a second molecule of water.  The reaction is therefore written:
   2H₂ + O₂ 2H₂O In the chemical reaction above, the number in front of the molecule (called a coefficient) indicates how many molecules participate in the reaction.  A simulation of the reaction can be viewed by clicking below (the atoms are represented as spheres in the animation: red = hydrogen, blue = oxygen):

(~61k animation opens in a new window)

        In order to write a correct chemical reaction, we must balance all of the atoms on the left side of the reaction with the atoms on the right side.  Let's look at another example.  Natural gas is primarily methane.  Methane (CH₄) is a molecule in which 4 hydrogen atoms are bonded to one carbon atom.  If you have a gas stove, lighting the stove causes the methane to react with oxygen in the atmosphere to release heat and the atoms recombine to form carbon dioxide and water vapor.  The unbalanced chemical reaction would be:
CH₄ + O₂ CO₂ + H₂O Look at the reaction atom by atom.  On the left side we find 1 carbon atom, and 1 on the right.  There are 4 hydrogen atoms on the left, but only 2 on the right.  Therefore, you know 2 water molecules must be formed.  Adding this coeffiecient we get:
CH₄ + O₂ CO₂ + 2H₂O  Now we have to balance the oxygen atoms.  On the left you find 2 atoms, on the right 4 (2 in the CO₂ molecule and 1 in each of 2 H₂O molecules).  Therefore we need to start with 4 oxygen atoms, or 2 molecules.  The balanced equation would then be:
CH₄ + 2O₂ CO₂ + 2H₂O Try balancing the following equations on your own (note: while you don't need to write the coefficient 1 in a chemical reaction, these examples will not work unless you input a 1 where needed):
1) Na + Cl₂NaCl

2) N₂ + H₂NH₃

3) Fe + O₂Fe₂O₃

4) Cu + AgNO₃Ag + Cu(NO₃)₂ (a Cu atom bonded to 2 NO₃ groups)

5) H₂SO₄ + NaCN HCN+ Na₂SO₄

The Mole and Molecular Weights         Up until this point we have been talking about atoms and molecules.  The problem with this approach is that atoms and molecules are very small things.  In a single drop of water for example, there are trillions and trillions of water molecules.  A reaction between a single molecule of hydrogen and a single molecule of oxygen, as we discussed above, would be undetectable.  Instead of talking about single molecules in science, we talk about groups of molecules.  You can think of it like buying eggs.  You don't go to the store and buy an egg - you buy a dozen.  Contained within that dozen are the individual eggs.  Its the same thing when we talk about molecules.  We don't talk about single units, we talk about groups.
        But even a dozen molecules is a tiny amount.  What we need is a big number - a huge number!  That number is the mole.  The mole is the scientific community's baker's dozen.  One mole equals 6.02 x 10²³ (also known as Avogadro's number).  A 6 followed by 23 zeros.  Now that's a pretty big number.  But that's all it is, a number.  You can't just have a mole, you have to have a mole of something.  A mole of atoms.  A mole of water molecules.  A mole of pennies (which would make you richer than you can imagine).  Why the mole?  As it turns out, the mole has some interesting properties.  One mole of hydrogen atoms (6.02 x 10²³ H atoms) weighs 1 g.  From the periodic table we know that an He atoms weighs 4 times as much as an H atom, so go figure, 1 mole of He atoms weighs 4 g.  In fact, one mole of any element is equal to the atomic mass of that element (in grams).
        Let's think about that for a second.  If we know the molar mass of an element, and we know how many elements make up a specific molecule, then you can calculate the molar mass of a compound by adding up the atomic weights.  Huh?  Take water for example.  How much does a mole of water weigh?  Well, one mole of water contains one mole of oxygen atoms and two moles of hydrogen atoms.  A mole of hydrogen weighs 1 g and a mole of oxygen weighs 16 g (look at the atomic mass in the periodic table).  So to calculate the weight of one mole of water:
(2 moles H * 1 g per mole) + (1 mole O * 16 g per mole) = 18 g One mole of water weighs 18 grams!
        The mole is also useful in chemical reactions.  Though you can't measure out an atom of hydrogen, you can measure out a mole.  Since the mole is just a constant number, the coefficients in a balanced chemical reaction give you the molar proportions of reactants and products.  In other words:
   2H₂ + O₂ 2H₂O tells us that: 2 H₂ molecules react with 1 O₂ molecule to form 2 H₂O molecules. It also tells us that:
2 moles of H₂ molecules react with 1 mole of O₂ molecules to form 2  moles of H₂O molecules.

          There are many different types of chemical reactions.  Chemists have classified the many different reactions into general categories.  The chemical reactions we will explore are a representation of the types of reactions found in each group.  There is a general description of the main reaction types and specific examples provided in the  selection boxes.

Synthesis Reaction (Combination Reaction)

In a synthesis reaction, two or more substances combine to form a new compound.  This type of

reaction is represented by the following equation.

A       +       B                          AB

          A and B represent the reacting elements or compounds while AB represents a compound as the product.

The following examples are representative of synthesis reactions. 

Aluminum and Bromine

Formation of Aluminum Bromide:  When Al is placed on the surface of liquid Br2 an exothermic reaction occurs. The Al is oxidized to Al3+ by the Br2, which is reduced to Br - ions. The ionic product, AlBr3, can be observed on the watch glass after the reaction.

Sodium and Chlorine

Formation of Sodium Chloride:  Molten sodium burns when it is put into a container of chlorine gas. In the reaction a sodium ion loses an electron to form a sodium cation and a chlorine atom simultaneously gains an electron to form a chloride anion. The product of the reaction is the ionic compound sodium chloride, which is the white solid observed.

Zinc and Oxygen

Formation of Zinc Oxide:  Oxidation is a loss of electrons and reduction is a gain of electrons. The oxidation of metallic Zn by O2 to form ZnO(s) is illustrated at the molecular level. The transfer of electrons from Zn to O2 is shown. Atoms can be observed to change as they are oxidized or reduced, respectively to their ionic forms.

Sodium and Potassium in Water

Formation of Sodium Hydroxide and Potassium Hydroxide:  When a small piece of Na is added to a solution containing an indicator, evidence of the reaction can be observed by the change in the color of the solution as NaOH is formed, by the melting of the Na and by the movement of the Na caused by formation of hydrogen gas. K is more reactive than Na as demonstrated by its reaction with water. This reaction produces enough heat to ignite the H2 produced.

 

Single-Replacement Reaction

         

          In a single-replacement reaction (displacement reaction) one element replaces a similar element

in the compound.  Single-replacement reactions can be represented by the following equations.

AB    +       C                          AC    +       B

Iron (III) Oxide and Aluminum Reaction 2

Thermite Reaction:  In the thermite reaction, Al reduces Fe2O3 to Fe in an extremely exothermic reaction in which Al is oxidized to Al2O3. The reaction produces enough heat to melt the iron. Because of the extreme heat produced in the thermite reaction, it is used industrially to weld iron.

Copper (II) Oxide and Carbon

Reduction of CuO:  When black carbon and black copper oxide are heated together the Cu2+ ions are reduced to metallic Cu and a gas is evolved. When the gas is collected in Ca(OH)2 a white precipitate of CaCO3 is formed. The reaction which occurs involves the reduction of Cu2+ ions by carbon which is oxidized to CO2.

Silver Nitrate and Copper

Formation of Silver Crystals:  When a copper wire is placed in a solution of AgNO3, the Cu reduces Ag+ to metallic Ag. At the same time, Cu is oxidized to Cu2+. As the reaction progresses Ag crystals can be seen to form on the Cu wire and the solution becomes blue as a result of the formation of Cu2+ ions.

Tin (II) Chloride and Zinc

Formation of Tin Crystals:  Oxidation-reduction chemistry of Sn and Zn. When acidified Sn(II)Cl2 is added to a beaker containing a piece of Zn, some of the Sn2+ reacts with H+ in the solution to produce H2 gas. Immediate changes can also be observed on the surface of the Zn as it quickly becomes coated with Sn crystals. After the reaction has progressed for a time needles of Sn can be observed on the surface of the Zn.

Double-Replacement Reaction

          In a double-replacement reaction, the ions of two compounds exchange places in an aqueous solution

to form two new compounds.  A double-replacement reaction can be represented by the following equation.

AB    +      CD                   AC   +      BD

Calcium carbonate and Sulfurous Acid

This marble statue has been eroded by acid rain. Marble is a material having CaCO3 as its primary component. Acids react with and dissolve the marble.  The acid comes from sulfur dioxide in the atmosphere combining with water to form sulfurous acid.

Lead (II) Nitrate and Potassium Iodide

An aqueous solution of Potassium Iodide is added to an aqueous solution of Lead (II) Nitrate forming lead (II) iodide.  The formation of a precipitate occurs when the cations of one reactant combines with the anions of the other reactant to form an insoluble or slightly insoluble compound.

Sodium chloride and Silver nitrate

An aqueous solution of Sodium Chloride is added to an aqueous solution of Silver Nitrate forming silver chloride.

Decomposition Reaction

In a decomposition reaction, single compound undergoes a reaction that produces two or more simpler

substances.  A decomposition reaction can be represented by the following equation.

AB                            A      +      B

Water into Hydrogen and Oxygen

Electrolysis of Water:  When a direct current is passed through water it decomposes to form oxygen and hydrogen. The volume of hydrogen gas produced at the negative electrode is twice the volume of the oxygen gas formed at the positive electrode. This indicates that water contains twice as many hydrogen atoms as oxygen atoms, which is an illustration of the law of constant composition.

Nitrogen Triiodide

Decomposition of Nitrogen Triiodide: Nitrogen triiodide is extremely unstable when it is dry. Touching it with a feather causes it to decompose explosively. The explosion occurs as chemical energy is released by the decomposition of nitrogen triiodide to N2 and I2. Violet iodine vapor can be observed after the explosion.

Combustion Reaction           

          In a combustion reaction, a substance combines with oxygen, releasing a large amount of energy in the form

of light and heat.  For organic compounds, such as hydrocarbons, the products of the combustion reaction are carbon dioxide and water.

                   CH₄  +      2 O₂                                      CO₂  +      2 H₂O

Hydrogen and Oxygen Reaction II

The combustion of hydrogen yields water vapor as a reaction product.  Three balloons of hydrogen and one balloon mixed with hydrogen and oxygen form an explosive mixture

Various Substances with Oxygen

Reactions with Oxygen. Magnesium, steel wool, white phosphorous, and sulfur are burned in oxygen. The resulting reactions are combination reactions in which two substances react to form one product. The products formed in these reactions are MgO, Fe2O3, P4O10 and SO2. All of these combustion reactions are very exothermic.

Phosphorous and Oxygen

The combustion of yellow phosphorus occurs in an oxygen atmosphere.  The main product of this reaction is phosphorus pentoxide.